A cylindrical block of length 0.4 m and area of cross-section 0.04 m^2 is placed coaxially on a thin

Q: Why do we use a differential equation instead of just $\Delta Q = \frac{KA \Delta T}{l} t$?

Because the temperature of the disc is not constant. As the disc heats up, the temperature difference between the top and bottom of the cylinder decreases, which means the rate of heat flow also decreases over time. A simple algebraic equation only works for steady-state constant heat flow.

Q: Why is the thermal conductivity of the disc assumed to be very high?

A very high thermal conductivity ensures that the heat distributes instantly throughout the disc, meaning the entire disc is at a uniform temperature $T$ at any given moment. This simplifies our calorimetry assumption.

Q: What does the negative sign in the integration of $\frac{1}{T_0 - T}$ mean?

It comes from the chain rule of integration. The derivative of $(T_0 - T)$ with respect to $T$ is $-1$. This negative sign is crucial and perfectly cancels out when we evaluate the limits, ensuring the calculated time is positive.

Q: Will the disc ever reach exactly 400 K?

Theoretically, it would take an infinite amount of time to reach exactly 400 K because the rate of heat flow approaches zero as the temperature difference approaches zero. It follows an exponential decay curve.

Q: What if the area of the disc was different from the cylinder?

The area $A$ in Fourier's law refers to the cross-sectional area through which heat conducts (the cylinder's area). The mass $m$ of the disc would still absorb this heat. The equation would remain the same, but you'd strictly use the cylinder's area for $A$ and the disc's actual mass for $m$.

Question

A cylindrical block of length $0.4 	ext{ m}$ and area of cross-section $0.04 	ext{ m}^2$ is placed coaxially on a thin metal disc of mass $0.4 	ext{ kg}$ and of the same cross-section. The upper face of the cylinder is maintained at a constant temperature of $400 	ext{ K}$ and the initial temperature of the disc is $300 	ext{ K}$. If the thermal conductivity of the material of the cylinder is $10 	ext{ W/mK}$ and the specific heat capacity of the material of the disc is $600 	ext{ J/kg-K}$, how long will it take for the temperature of the disc to increase to $350 	ext{ K}$? Assume, for purposes of calculation, the thermal conductivity of the disc to be very high and the system to be thermally insulated except for the upper face of the cylinder.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a solid cylinder resting perfectly on a thin metal disc. The top of this cylinder is maintained at a blazing hot, constant temperature of $T_0 = 400 	ext{ K}$. The disc at the bottom starts at a cooler initial temperature of $T_i = 300 	ext{ K}$.

Because the sides of the cylinder and the disc are perfectly insulated, heat has only one path to travel: straight down through the cylinder and into the disc. This means that every single joule of heat that conducts through the cylinder is entirely absorbed by the disc, raising its temperature.

### The Master Equation

We have two distinct thermal processes happening simultaneously. First, heat conducts through the cylinder. According to Fourier's Law of Heat Conduction, the rate of heat flow at any instant when the disc is at temperature $T$ is given by:

$$ \frac{dQ}{dt} = \frac{KA}{l}(T_0 - T) $$

Second, the disc absorbs this exact same heat. According to the principle of calorimetry, the rate at which the disc absorbs heat to raise its temperature is:

$$ \frac{dQ}{dt} = ms \frac{dT}{dt} $$

Since the heat conducted equals the heat absorbed, we can equate these two rates to form our master differential equation:

$$ ms \frac{dT}{dt} = \frac{KA}{l}(T_0 - T) $$

### Setting up the Integration

To find the total time $t$ required for the disc to reach a final temperature of $T_f = 350 	ext{ K}$, we need to separate the variables and integrate. We group all the temperature terms on one side and the time terms on the other:

$$ \frac{dT}{T_0 - T} = \frac{KA}{msl} dt $$

Now, we integrate both sides from the initial state ($t = 0, T = 300 	ext{ K}$) to the final state ($t = t, T = 350 	ext{ K}$):

$$ \int_{300}^{350} \frac{dT}{400 - T} = \int_{0}^{t} \frac{KA}{msl} dt $$

**Watch out for the minus sign!** The integral of $\frac{1}{400 - T}$ is $-\ln(400 - T)$. Applying the limits:

$$ \left[ -\ln(400 - T) ight]_{300}^{350} = \frac{KA}{msl} t $$

$$ -\ln(400 - 350) + \ln(400 - 300) = \frac{KA}{msl} t $$

$$ \ln(100) - \ln(50) = \frac{KA}{msl} t $$

Using the property of logarithms, $\ln(a) - \ln(b) = \ln(\frac{a}{b})$, this beautifully simplifies to:

$$ \ln(2) = \frac{KA}{msl} t $$

### Final Calculation

Now, we just need to rearrange for $t$ and plug in the given physical constants: $m = 0.4 	ext{ kg}$, $s = 600 	ext{ J/kg-K}$, $l = 0.4 	ext{ m}$, $K = 10 	ext{ W/mK}$, and $A = 0.04 	ext{ m}^2$.

$$ t = \frac{msl}{KA} \ln(2) $$

$$ t = \frac{(0.4)(600)(0.4)}{(10)(0.04)} \ln(2) $$

Don't rush the calculation here. Cancel out the decimals carefully. The denominator becomes $0.4$, which perfectly cancels with one of the $0.4$s in the numerator:

$$ t = 600 	imes 0.4 	imes \ln(2) $$

$$ t = 240 \ln(2) $$

Finally, substituting the approximate value of $\ln(2) \approx 0.693$:

$$ t = 240 	imes 0.693 = 166.32 	ext{ s} $$

And there we have it! It takes exactly **166.32 seconds** for the disc to heat up to 350 K under these perfectly insulated conditions.

Enter Numerical Value:

Visualized Solution

$Visualizing the Setup$

$Rate of Heat Flow$

$Forming the Differential Equation$

$Setting up the Integration$

$Evaluating the Integral$

$Substituting the Values$

$Final Calculation$

$The Way Forward$

The Sigma Insight: Heat Transfer

Analyzing the Setup

The Master Equation

Setting up the Integration

Final Calculation

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Rate of Heat Flow

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Forming the Differential Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Setting up the Integration

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluating the Integral

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting the Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward